Contoh Soal 10 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 46.&\textrm{Turunan kedua dari}\: \: f(x)=x^{3}-\sin 3x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&6x^{2}+9\sin 3x\\ \textrm{b}.&3x^{2}+6\sin 3x\\ \textrm{c}.&3x-9\sin 3x\\ \color{red}\textrm{d}.&6x+9\sin 3x\\ \textrm{e}.&9x-6\sin 3x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}f(x)&=x^{3}-\sin 3x\\ f'(x)&=3x^{2}-3\cos 3x\\ f''(x)&=6x+9\sin 3x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Diketahui fungsi}\: \: g(x)=\displaystyle \frac{1-\cos x}{\sin x}\: . \textrm{Nilai}\\ &\textrm{turunan kedua saat}\: \: x=\displaystyle \frac{\pi}{4}\: \: \textrm{adalah}\: .... \\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2}+4\\ \textrm{b}.&2\sqrt{2}-3\\ \textrm{c}.&2\sqrt{2}+3\\ \color{red}\textrm{d}.&3\sqrt{2}-4\\ \textrm{e}.&3\sqrt{2}+4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}g(x)&=\displaystyle \frac{1-\cos x}{\sin x}\\ g'(x)&=\displaystyle \frac{\sin x(\sin x)-\cos x(1-\cos x)}{\sin ^{2}x}\\ &=\displaystyle \frac{\sin ^{2}x-\cos x+\cos ^{2}x}{\sin ^{2}x}\\ &=\displaystyle \frac{1-\cos x}{\sin ^{2}x}\\ g''(x)&=\displaystyle \frac{\sin x(\sin ^{2}x)-2\sin x\cos x(1-\cos x)}{\sin ^{4}x}\\ &=\displaystyle \frac{\sin x(\sin ^{2}x)-\sin 2x(1-\cos x)}{\sin ^{4}x}\\ &=\color{red}\displaystyle \frac{\sin \displaystyle \frac{\pi }{4}(\sin ^{2}\displaystyle \frac{\pi }{4})-\sin 2\displaystyle \frac{\pi }{4}(1-\cos \displaystyle \frac{\pi }{4})}{\sin ^{4}\displaystyle \frac{\pi }{4}}\\ &=\color{black}\displaystyle \frac{\left ( \displaystyle \frac{1}{\sqrt{2}} \right )\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{2}-1.\left ( 1-\left ( \displaystyle \frac{1}{\sqrt{2}} \right ) \right )}{\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{4}}\\ &=\color{black}\displaystyle \frac{\displaystyle \frac{1}{2}\displaystyle \frac{1}{\sqrt{2}}-1+\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \frac{1}{4}}\times \displaystyle \frac{4}{4}\\ &=\displaystyle \frac{\displaystyle \frac{2}{\sqrt{2}}-4+\frac{4}{\sqrt{2}}}{1}\\ &=\displaystyle \frac{6}{\sqrt{2}}-4=3\sqrt{2}-4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Turunan kedua fungsi}\: \: f(x)=\sin ^{2}x-\cos ^{2}x\\ &\textrm{adalah}\: \: f''(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&6\sin 2x\\ \color{red}\textrm{b}.&4\cos 2x\\ \textrm{c}.&2\cos 2x\\ \textrm{d}.&-2\cos 2x\\ \textrm{e}.&-4\cos 2x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}f(x)&=\sin ^{2}x-\cos ^{2}x\\ f'(x)&=2\sin x\cos x-2\cos x(-\sin x)\\ &=2\sin x\cos x+2\sin x\cos x\\ &=2(2\sin x\cos x)=\color{black}2\sin 2x\\ f''(x)&=\color{red}2.2\cos 2x=4\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Diketahui}\: \: f(x)=\sqrt{\sin x}\: .\: \textrm{Jika}\: \: f''(x)\\ &\textrm{adalah turunan keduafungsi}\: \: f,\: \textrm{maka}\\ &\textrm{nilai dari}\: \: f''\left ( \displaystyle \frac{\pi }{2} \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\displaystyle \frac{1}{2}\\ \textrm{b}.&-\displaystyle \frac{1}{4}\\ \textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}f(x)&=\color{black}\sqrt{\sin x}=\sin ^{\frac{1}{2}}x\\ f'(x)&=\displaystyle \frac{1}{2}\sin ^{-\frac{1}{2}}x.\cos x=\displaystyle \frac{\cos x}{2\sin ^{\frac{1}{2}}x}\\ f''(x)&=\color{red}\displaystyle \frac{-\sin x\left ( 2\sin ^{\frac{1}{2}}x \right )-\cos x\left ( 2.\displaystyle \frac{1}{2}\sin ^{-\frac{1}{2}}x.\cos x \right )}{4\sin x}\\ &=\displaystyle \frac{-2\sin x\sqrt{\sin x}-\displaystyle \frac{\cos ^{2}x}{\sqrt{\sin x}}}{4\sin x}\\ f''\left ( \displaystyle \frac{\pi }{2} \right )&=\color{black}\displaystyle \frac{-2\sin \displaystyle \frac{\pi }{2}.\sqrt{\sin \displaystyle \frac{\pi }{2}}-\displaystyle \frac{\cos ^{2}\displaystyle \frac{\pi }{2}}{\sin \displaystyle \frac{\pi }{2}}}{4\sin \displaystyle \frac{\pi }{2}}\\ &=\displaystyle \frac{-2.1.1-0}{4.1}=-\frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \: f(x)=\tan ^{2}(3x-2) \: \: \textrm{maka}\: \: f''(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &-18\sec ^{4}(3x-2)\\ \textrm{b}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{2}(3x-2)\\ \color{red}\textrm{c}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{4}(3x-2)\\ \textrm{d}.&18\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+36\sec ^{4}(3x-2)\\ \textrm{e}.&18\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{4}(3x-2) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\tan ^{2}(3x-2)\\ f'(x)&=2\tan (3x-2)\sec ^{2}(3x-2)(3)\\ &=6\tan (3x-2)\sec ^{2}(3x-2)\\ f''(x)&=6\sec ^{2}(3x-2).(3)\sec ^{2}(3x-2)\\ &+6\tan (3x-2).2\sec (3x-2).\sec (3x-2)\tan (3x-2)(3)\\ &=18\sec ^{4}(3x-2)\\ &+36\tan ^{2}(3x-2)\sec ^{2}(3x-2) \end{aligned} \end{array}$